How do we know that every function of a DE we "solved" satifies the equation? Suppose we have a D.E.  $\frac{dy}{dx}=f(y,x)$, which we solve by taking functions of both sides, dividing integrateing etc.
How do we know that every function of a  DE we solved satifies the equation? And how do we know that we found all solutions?
For example, suppose we use some function on both sides which is not bijective on the range of expression then we surley loose some solution right? How would one keep track of which solutions one looses or even introduce using different kinds of operations?
 A: With ODEs it's as with all equations or systems of equations: An "equation" (or system, incl. extra conditions, etc.)  $\Phi(x)=0$ defined for elements $x$ in some base set $X$  implicitly defines a solution set
$$S:=\bigl\{x\in X\,\bigm|\,\Phi(x)=0\bigr\}\subset X\ .$$
It follows that for any $x\in X$ it is easy to test whether $x\in S$; but we do not have an a priori overview over $S$. The task is to obtain an explicit presentation of $S$ in the form of a finite list $S=\{x_1,x_2,\ldots, x_r\}$, or a parametric representation which produces all elements of $S$ in a bijective way.
Confronted with a "new" ODE one usually applies the procedures you describe in the form of a chain
$$\Phi(x)=0\quad\Rightarrow\ldots\Rightarrow\ldots\quad\Rightarrow x\in Q\ ,\tag{1}$$
where now $Q$ is an explicitly given set of functions, say the set $y(t)=Ce^t$, where $C\in{\mathbb R}$ is arbitrary. But herewith we only have proven $S\subset Q$, and now have to test for each $x\in Q$ whether it is actually a solution to the original equation. By the way: There may also be a small leak in the chain $(1)$, and we have lost some special solutions which then do not appear in $Q$. As an example consider the ODE $x'=x^{2/3}$ where the special solution $x(t)\equiv0$ tends to get lost in the usual procedure.
It is a different matter if your ODE is of a type for which you have a general theory that guarantees "a three-dimensional vector space of solutions" or "exactly one solution". An example: The ODE $x'''-2x''+x=0$ is a linear homogeneous equation of third order. The theory says that the solution set $S$ is a vector space of dimension $3$. If you find, by whichever means, three linearly independent solutions $x_1$, $x_2$, $x_3$ then you can say right away that the general solution is given by $t\mapsto C_1x_1(t)+\ldots+C_3x_3(t)$. Similarly in the case of an IVP satisfying the regulatory assumptions. If in the course of doing $(1)$ you obtain $Q=\{x_*\}$, then you can say that $x_*$ is the solution of the given IVP without performing further tests.
